Differential forms are important concepts in mathematics and
have ready
applications in physics, but their nature is not intuitive. In contrast
the concept of vectors and vector fields can be easily grasped. The purpose
of this site is to explain the nature of differential forms, both the
formal definitions and how they are used. The first step is to assert that
differential forms are related to vectors in a very subtle way. They are, in
fact, dual to vectors. The meaning of this will be explained below.

First consider a vector space. Roughly, a vector space is a set of entities
such that the sum of any vectors is also a vector and the result of
multiplying a vector by a scalar is also a vector. A vector space could
be the set of arrows in space having direction and magnitude. But the set
of solutions to a linear differential equations also constitutes a vector
space because the sum of any two solutions is also a solution, as is a
multiple of any solution. A remarkable fact is that any vector space
has a basis, meaning that there exists a subset of vectors in the space
such that any vector in the space can be represented as a linear combination
of members of the basis. Thus the vector space can be considered to be
ordered n-tuples of scalars involved in the linear combinations.

Formally the vector space is a set of four things, *(V,k,+,*) where
V is the set of vectors, K is the field of scalars involved in creating
multiples of vectors, + is the function involved in adding two vectors and
* is the binary function involved in multiplying a vector by a scalar.

Now consider the set of functions defined on V that have values in K; i.e.,
f:V → K. Within this set of functions there are linear functions such
that if V1 and V2 are vectors then
f*(V1 + V2) = f*(V1) + f*(V2) and
for any scalar k and any vector V, f*(k*V) = kf*(V). Such linear functions
on vector spaces are called linear functionals.

The sum of two linear functionals f1 and f2 can be
defined by *(f1+f2)*(V) = f1*(V)
+ f2*(V)
where the addition on the right is addition in the field of scalars, K.
Likewise for any linear function f and any scalar k their product can
be defined as *(kf)*(V) = kf*(V), again where the multiplication on the
right is the multiplication of the scalar field. What these definitions of
addition of linear functionals and scalar multiplication of linear functions
is that the set of linear functionals, with the scalar field and the
definitions of addition and multiplication constitute a vector space. The
vector space of linear functionals over V is said to be dual to the
vector space involving V. For vector spaces with finite bases the dual
spaces are not very exotic; they are essentially the same as the original
spaces. If the original space is a finite dimensional column vector then
the dual space to it is a space of row vectors of the same dimension.
There some infinite dimensional vector spaces that have dual spaces that
are different in nature from the original space.

Differential forms are the dual spaces to the spaces of vector fields over
Euclidean spaces. Vector fields over some space X are a bit more complex
than vector spaces of n-tuples. At each point in the space X there is
a vector, say F*(x,y,z). This is equivalent to functions of the
form f:X→V; i.e., functions which map every point of the space X into
the vector space V.

The vector space of linear functionals over vector fields will have a
basis. Suppose the basis for the vector space V is denoted as
*(ix,iy,iz), where ix is a unit
vector in the positive x direction and so forth. Suppose a basis for the
vector space of linear functionals over V is denoted as
*(dx,dy,dz).
The symbol dx denotes the linear functional which selects the
x-component of any vector v; i.e.,
if v = vxix+vyiy+vziz
then dx*(v)=vx. Likewise for dy and
dz. The application of these linear functionals to the
basis vectors of V give:

Thus the basis of the vector space of linear functionals over the vector
space V is conjugate to the basis of V.
*(dx,dy,dz) is a basis, any linear functional
w may be represented as wxdx+wydy+wzdz.
The value of w*(v) for any v=vxix+vyiy+vziz
is wxvx+wyvy+wzvz.

If the basis for the linear functionals is written as *(*dx, *dy, *dz) instead
of *(dx,dy,dz) nothing has changed. In this
slightly modified notation any linear functional w can be represented as
w = wx*dx + wy*dy + wz*dz.

A vector field is a function over a region of three dimensional
space that gives at each point a vector. For a point P=*(x,y,z) let the
vector be v*(P) = vx*(P)ix+vy*(P)iy+vz*(P)iz.
The corresponding dual entity is a function over the region of three
dimensional space which gives a linear functional at each point; i.e.,
w*(P) = wx*(P)*dx+wy*(P)*dy+wz*(P)*dz. This
is called a 1-form, a special case of a differential form. Thus a 1-form is
a field of linear functionals.

The next task is to define k-forms where k can be 0, 2 or 3 as well as 1.
However, at this point it should be noted that there are alternative
approaches to defining differential forms. One approach defines differential forms
of the various dimensions in terms of symbolic expressions. Let P denote
*(x, y, z). Then:

0-forms are scalar functions of the form f*(P)

1-forms are expressions of the form f*(P)*dx+g*(P)*dy+h*(P)*dz

2-forms are expressions of the form f*(P)*dx*dy+g*(P)*dy*dz+h*(P)*dz*dx

3-forms are expressions of the form f*(P)*dx*dy*dz.

A wedge *(^) product of differential forms can be defined for these
symbolic expressions. Also the differenttiation of a k-form to produce
a *(k+1)-form is defined for these symbolic expressions. This approach
is consistent but the lack of a geometric interpretation of the symbolic
expressions destroys motivation to learn this topic.

Symbolically 2-forms involve expressions of the form *dx*dy. How can the
expression *dx*dy be given some concrete, geometric meaning? Williamson,
Crowell and Trotter in their Calculus of Vector Functions provide
an answer. They assert that *dx*dy is a linear function of order pairs of
vectors such that for the vector pair *(v,w), *dx*dy*(v,w) is equal to the
area of the parallelogram formed by the projections of v and w onto the
xy-plane, as shown in the diagram below.

This sounds excessively abstruse but computationally it is quite simple,
being the value of the determinant of a 2x2 matrix; i.e.,

| *dx*(v)

*dx*(w) |

| *dy*(v)

*dy*(w) |

which is equal to

| v1

w1 |

| v2

w2 |

The value of the determinant is just v1w2 - v2w1.

The meaning of *dy*dz and *dz*dx is given by analogous expressions so that
the general definition of *dxi*dxj*(v1,v2)
is given by:

| *dxi*(v1)

*dxi*(v2) |

| *dxj*(v1)

*dxj*(v2) |

which is equal to

| v1i

v2i |

| v1j

v2j |

It immediately follows from the properties of the determinant that if
i=j the determinant is equal to zero. Thus, *dxi*dxi
equals the zero functional for all values of i. It also follows that
*dxj*dxi = -*dxi*dxj.

This approach can be generalized to give the definition of other
basic k-forms.

General Definition of Differential Forms

Basic Forms

A k-form *dxi1*dxi2...*dxik1
is the linear function over k-tuples of vectors *(v1,v2,...,v1)
equal to the determinant of the matrix whose general term is
*dxi*(vj); i.e., the element in the i-th row,
j-th column is the i-th component of the j-th vector, vji.

General Forms

Roughly a k-form is the formal sum of functions over the space f*(P) times the
basic k-forms. More properly a k-form is a vector whose components are
functions over the space and the basis of these vectors is the basic
k-forms.

With this founding of differential forms in geometry and analysis the
formal expression approach to differential forms becomes a valuable and
powerful method of computing wedge products and derivatives of
differential forms. It is all very beautifully tied together.

The Wedge Product of Differential Forms

From the definition in terms of a determinate it is seen than any
n-tuple of differentials, such as *dxi*dxj or
*dxi*dxj*dxk, is the zero linear functional if any index is
repeated. Furthermore if any two indices are interchanged the sign of
the n-tuple is reversed. Thus *dx*dx=*dy*dy=*dz*dz=0 and *dy*dx=-*dx*dy and so on.
The wedge product of two differential forms can be computed by creating
their formal product and reducing it using the above relations.

For example, consider the product of y*dx+x2*dy with
z*dx+*dy-x*dz. The formal product is given by:

*(f ^ ω) = fω

The Derivative of Differential Forms

The operation of taking the derivative of differential forms is defined
recursively.
The derivative of a 0-form, a scalar function f*(x,y,z) over the space,
is defined as

df = *(∂f/∂*dx)*dx + *(∂f/∂*dy)*dy + *(∂f/∂*dz)*dz.

This is called the gradient of f and is the base case for the recursive
definition of the derivative. The derivative of the sum and the
wedge product of two differential 1-forms, φ and ω, is given by the rules:

d*(φ + ω) = dφ + dω

d*(φ ^ ω)=*(dφ ^ ω) - *(φ ^ dω)

From these rules we quickly find that d*(dω)=0 for any differential
1-form ω because it can be consider the product of the constant function
equal to 1 and itself. Thus

d*(dω) = d*(1dω) = d*(1^ω) =
d*(1)^ω - *(1^d*(dω)) = - d*(dω)

but d*(dω) = - d*(dω) can only be true if d*(dω)=0.
In particular, since *dx=d*(x), d*(*dx)=0 and likewise d*(*dy)=0 and d*(*dz)=0.

Here is how these rules would be applied to get the derivative of
ω = xz*dx - y*dy.

dω = d*(xz*dx - y*dy) = d*(xz*dx) - d*(y*dy)

=
*(d*(xz)^*dx - xzd*(*dx)) - *(*dy^*dy - *(y^d*(*dy))

= *(z*dx+x*dz)^*dx - 0) - *(0 - 0) = x*dz*dx = - x*dx*dz.

Although the definition of the derivative of the wedge product of 1-forms
is as given above this is not the general formula for the derivative
of the wedge product of differential forms of arbitrary degree. In particular,
it would not hold for the case of 0-forms; i.e., scalar-valued functions.
For scalar functions f and g:

d*(f^g) = d*(fg) = *(df)g + f*(dg) = df^g + f^dg

Also for the wedge product of a 0-form f and a 1-form ω we have

d*(f^ω) = d*(fω) = df^ω + fdω.

Although it is not obvious from these special cases the proper
definition of the derivative of the wedge product of an i-form φ and
a j-form ω is:

d*(φ^ω) = *(dφ)^ω + *(-1)iφ^*(dω)
which is the same as
d*(φ^ω) = *(dφ)^ω + *(-1)deg*(φ)φ^*(dω)

General Commutativity Relation for the Wedge Product

Although it was found that for two 1-forms, φ and ω
φ^ω = -ω^φ, this is not the general
relation for differential forms of arbitrary degree. For an i-form
φ and a j-form ω the commutativity relation is:

φ^ω = *(-1)ijω^φ,
which is the same as
φ^ω = *(-1)deg*(φ)deg*(ω)ω^φ.

This formula may be explained by considering the process involved in
interchanging an i-form of the form

*dxp1*dxp2...*dxpiwith a j-form of the form
*dxq1*dxq2...*dxqj.

The terms of the j-form have to be passed through the i-form one by one. Each
term of the j-form has to be interchanged with successive terms of the
i-form. One cannot jump *dxq1 to the first position
because that would change the order of the terms of the terms of the i-form.
Since each interchange reverses the sign and it requires i interchanges
to pass one term of the j-form through the terms of the i-form the result
is:

Since there are j terms that must be passed through the i-form the final
sign will be *(*(-1)i)j which is equal to
*(-1)ij. The same sign would apply to each wedge product of
the product terms in the wedge product of a general i-form with a general
j-form.

General Formula for the Derivative of the Wedge Product of Two Differential Forms

Let us start with the case of the product of two simple forms. Let one
differential form be fφ = f*dxi1...*dxip and
and the other gω = g*dxj1......*dxjq.
Thus *(fφ)^*(gω) = fg*(φ^ω).
The derivative is therefore

The problem is the last term. To put it in proper form we must pass
each of the *dxj through φ by a sequence of interchanges.
This process results in a change of sign for each interchange. The final
result is then multiplied by *(-1)deg*(φ). The final result is
thus: